handouts/src/Advanced/Lambda Calculus/parts/03 numbers.tex

\section{Numbers}

Since the only objects we have in $\lm$ calculus are functions, it's natural to think of quantities as \textit{adverbs} (once, twice, thrice,...) rather than \textit{nouns} (one, two, three ...) 

\vspace{1ex}

We'll start with zero. If our numbers are \textit{once,} \textit{twice,} and \textit{twice}, it may make sense to make zero \textit{don't}. \par
Here's our \textit{don't} function: given a function and an input, don't apply the function to the input.
$$
	0 = \lm fa.a
$$
If you look closely, you'll find that $0$ is equivalent to the false function $F$.

\problem{}
Write $1$, $2$, and $3$. We will call these \textit{Church numerals}.\hspace{-0.5ex}\footnotemark{} \\
\note{\textit{Note:} This problem read aloud is \say{Define \textit{once}, \textit{twice}, and \textit{thrice}.}}

\footnotetext{after Alonzo Church, the inventor of lambda calculus and these numerals. He was Alan Turing's thesis advisor.}

\begin{solution}
	$1$ calls a function once on its argument: \par
	$1 = \lm fa . (f~a)$.

	\vspace{1ex}

	Naturally, \par
	$2 = \lm fa . [~f~(f~a)~]$ \par
	$3 = \lm fa . [~f~(f~(f~a))~]$

	\vspace{1ex}

	The round parentheses are \textit{essential}. Our lambda calculus is left-associative! \par
	Also, note that zero is false and one is the (two-variable) identity.
\end{solution}

\vfill

\problem{}
What is $(4~I)~\star$?

\vfill

\problem{}
What is $(3~NOT~T)$? \par
How about $(8~NOT~F)$?

\vfill

\pagebreak

\problem{}
Peano's axioms state that we only need a zero element and a \say{successor} operation to
build the natural numbers. We've already defined zero.
Now, create a successor operation so that $1 \coloneqq S(0)$, $2 \coloneqq S(1)$, and so on. \par
\hint{A good signature for this function is $\lm nfa$, or more clearly $\lm n.\lm fa$. Do you see why?}

\begin{solution}
	$S = \lm n. [\lm fa . f~(n~f~a)] = \lm nfa . [f~(n~f~a)]$

	\vspace{1ex}

	Do $f$ $n$ times, then do $f$ one more time.
\end{solution}

\vfill

\problem{}
Verify that $S(0) = 1$ and $S(1) = 2$.

\vfill
\pagebreak


Assume that only Church numerals will be passed to the functions in the following problems. \par
We make no promises about their output if they're given anything else.

\problem{}
Define a function ADD that adds two Church numerals.

\begin{solution}
	$\text{ADD} = \lm mn . (m~S~n) = \lm mn . (n~S~m)$
\end{solution}

\begin{instructornote}
	Defining \say{equivalence} is a bit tricky. The solution above illustrates the problem pretty well.

	\note{Note: The notions of \say{extensional} and \say{intentional} equivalence may be interesting in this context. Do some reading on your own.
	}

	\vspace{4ex}

	These two definitions of ADD are equivalent if we apply them to Church numerals. If we were to apply these two versions of ADD to functions that behave in a different way, we'll most likely get two different results! \\
	As a simple example, try applying both versions of ADD to the Kestrel and the Kite.

	\vspace{1ex}

	To compare functions that aren't $\alpha$-equivalent, we'll need to restrict our domain to functions of a certain form, saying that two functions are equivalent over a certain domain. \\
\end{instructornote}
\vfill

\problem{}
Design a function MULT that multiplies two numbers. \par
\hint{The easy solution uses ADD, the elegant one doesn't. Find both!}

\begin{solution}
	$\text{MULT} = \lm mn . [m~(\text{ADD}~n)~m]$

	$\text{MULT} = \lm mnf . [m~(n~f)]$
\end{solution}

\vfill
\pagebreak


\problem{}
Define the functions $Z$ and $NZ$. $Z$ should reduce to $T$ if its input was zero, and $F$ if it wasn't. \par
$NZ$ does the opposite. $Z$ and $NZ$ should look fairly similar.

\vspace{1ex}

\begin{solution}
	$Z\phantom{N} = \lm n .[n~(\lm a.F)~T]$ \par
	$NZ = \lm n .[n~(\lm a.T)~F]$
\end{solution}

\vfill

\problem{}
Design an expression PAIR that constructs two-value tuples. \par
For example, say $A = \text{PAIR}~1~2$. Then, \par
$(A~T)$ should reduce to $1$ and $(A~F)$ should reduce to $2$.

\begin{solution}
	$\text{PAIR} = \lm ab . \lm i . (i~a~b) = \lm abi.iab$
\end{solution}

\vfill
From now on, I'll write (PAIR $A$ $B$) as $\langle A,B \rangle$. \par
Like currying, this is only notation. The underlying logic remains the same.

\pagebreak

\problem{}<shiftadd>
Write a function $H$, which we'll call \say{shift and add.} \par
It does exactly what it says on the tin:

\vspace{1ex}

Given an input pair, it should shift its second argument left, then add one. \par
$H~\langle 0, 1 \rangle$ should reduce to $\langle 1, 2\rangle$ \par
$H~\langle 1, 2 \rangle$ should reduce to $\langle 2, 3\rangle$ \par
$H~\langle 10, 4 \rangle$ should reduce to $\langle 4, 5\rangle$

\begin{solution}
	$H = \lm p . \Bigl\langle~(p~F)~,~S(p~F)~\Bigr\rangle$

	\vspace{1ex}

	Note that $H~\langle 0, 0 \rangle$ reduces to $\langle 0, 1 \rangle$
\end{solution}


\vfill

\problem{}<decrement>
Design a function $D$ that un-does $S$. That means \par
$D(1) = 0$, $D(2) = 1$, etc. $D(0)$ should be zero. \par
\hint{$H$ will help you make an elegant solution.}

\begin{solution}
	$D = \lm n . \Bigl[(~n~H~\langle 0, 0 \rangle~)~T\Bigr]$
\end{solution}

\vfill
\pagebreak